Vertex of the parabola $2 y^{2} + 3 y + 4 x - 2 = 0$ is

(\frac{25}{32}, \frac{- 7}{4})

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(\frac{25}{32}, \frac{- 3}{4})

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(\frac{15}{32}, \frac{7}{4})

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(\frac{17}{32}, \frac{- 3}{4})

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Solution

The correct option is B $(\frac{25}{32}, \frac{- 3}{4})$
$\begin{matrix} 2 y^{2} + 3 y - 2 + 4 x = 0 \Rightarrow y^{2} + \frac{3}{2} y - 1 + 2 x = 0 \Rightarrow y^{2} + 2 \cdot y \cdot \frac{3}{4} + \frac{9}{16} - \frac{9}{16} - 1 + 2 x = 0 \Rightarrow {(y + \frac{3}{4})}^{2} - \frac{25}{16} = - 2 x \Rightarrow {(y + \frac{3}{4})}^{2} = \frac{25}{16} = - 2 x \Rightarrow {(y + \frac{3}{4})}^{2} = (\frac{25}{16} - 2 x) \Rightarrow {(y + \frac{3}{4})}^{2} = - 2 (\frac{- 25}{32} + x) H e n c e, v e r t e x i s (\frac{25}{32}, \frac{- 3}{4}) \end{matrix}$

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